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Musya8 [376]
2 years ago
9

Samantha is making a vegetable soup that contains carrots, beans, water, salt, pepper, bits of ham, and onions. the soup has to

cook for 4 hours. while the soup is cooking, what will happen to some of the minerals that are in its ingredients?
Chemistry
1 answer:
murzikaleks [220]2 years ago
8 0
<span>While the soup is cooking, some of the minerals that are in its ingredients will flow out and mixed or react with the compounds in the water. It follows the law of osmosis where the flow is from a higher concentration to a lower concentration. That is why, at the end of cooking time, the water which was originally bland in taste, will taste like meat and veggie broth.</span>
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Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base
Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0
2 years ago
The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p
zysi [14]

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})  (in terms of mass)

or, q=\Delta H_{fusion}\times Moles     (in terms of moles)

Now we have to calculate the value of q.

q=6020J/mole\times 1Mole=6020J

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of \Delta e=0

Now we have to calculate the value of w.

Formula used :    \Delta e=q+w

where, q is heat required, w is work done and \Delta e is internal energy.

Now put all the given values in above formula, we get

0=6020J+w

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0
2 years ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
2 years ago
Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O
oee [108]

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

<u>To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺</u>:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the <u>number of Cr atoms</u> on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, <u>6 electrons are gained</u> by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to <u>balance the total charge</u> on the reactant and product side,<u> 14 H⁺ is added on the reactant side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to <u>balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

<u>Therefore, the correct balanced reduction half-reaction is:</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

3 0
2 years ago
Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
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